Function f ( x ) = ( x - 2 ) | x - 3 | is monotonically increasin… - Vidyadip

MCQ

Function $f\left( x \right) = \left( {x - 2} \right)\left| {x - 3} \right|$ is monotonically increasing in

✓
$\left( { - \infty ,\frac{5}{2}} \right) \cup \left( {3,\infty } \right)$
B
$\left( {\frac{5}{2},\infty } \right)$
C
$\left( {2,\infty } \right)$
D
$\left( { - \infty ,3} \right)$

✓

Answer

Correct option: A.

$\left( { - \infty ,\frac{5}{2}} \right) \cup \left( {3,\infty } \right)$

a
Given : $f(x)=(x-1)|(x-2)(x-3)|$

at $x < 0 \quad f(x)=(x-1)(-x-2)(-x-3)$

$=(x-1)(x+2)(x+3)$

$(x-1)\left(x^2+3 x+2 x+6\right)$

$=(x-1)\left(x^2+5 x+6\right)$

$=x^3+5 x^2-1 x^2-6$

$f(x)=x^3+4 x^2-6$

Differentiating w.r.t $x$

$f^1(x)=3 x^2-12 x=11$

Evaluate $f ^1( x )=0$

Given $f(x)=(-x)$

Given: $f(x)=(x-1)|(x-2)(x-3)|$

Finding critical points for $f(x)$

$f(x)=(-x+2)(-x+3)(x-1)$

$f^1(x)=3 x^2-12 x-11$

Evaluate $f ^{ l }( x )=0$

$3 x^2-12 x=11=0$

Therefore for $x < 2$

$x =\frac{6-\sqrt{3}}{3}$

$x=\alpha-\frac{1}{\sqrt{3}}$

Therefore the intervals where the function is increasing is

Therefore $\left(2-\frac{1}{\sqrt{3}}, 2\right)$ is the interval where the function is decreasing.

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